(23 :4).2(x+1)=64
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\(\widehat{M_3}+\widehat{N_3}=180^0\) Ma \(\widehat{N_3}+\widehat{N_1}=180^0\) va \(\widehat{M_2}=\widehat{M_3}\)
suy ra \(\widehat{M_2}=\widehat{N_1}\Rightarrow a//b\)
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ok nhưng mk làm bn phải k cho mk nha
trưới tiên mk có 1 câu hỏi : Bài đâu mà làm ?
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Bài làm:
Ta có: \(a^2.\left(a+1\right)=36\)
\(\Leftrightarrow a^3+a^2-36=0\)
\(\Leftrightarrow\left(a^3-3a^2\right)+\left(4a^2-12a\right)+\left(12a-36\right)=0\)
\(\Leftrightarrow a^2\left(a-3\right)+4a\left(a-3\right)+12\left(a-3\right)=0\)
\(\Leftrightarrow\left(a-3\right)\left(a^2+4a+12\right)=0\)
Mà \(a^2+4a+12=\left(a+2\right)^2+8>0\)
\(\Rightarrow a-3=0\Rightarrow a=3\)
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Đặt mỗi tòa nhà theo tỉ lệ 2 ; 3 ; 4 là \(a;b;c\left(a;b;c>0\right)\)
Vì a;b;c theo tỉ lệ 2 ; 3 ; 4 nên \(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}\). Mà tổng 3 tòa nhà có 117 căn nên
\(a+b+c=117\). Áp dụng tính chất dãy tỉ số bằng nhau, ta có :
\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=\frac{a+b+c}{2+3+4}=\frac{117}{9}=13\)
\(\Leftrightarrow a=13.2=26;b=13.3=39;c=13.4=52\)
Vậy số căn hộ mỗi tòa nhàn lần lượt là : 26 ; 39 và 52
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ta có
\(\left(x-1\right)\&\left(8-x\right)\in B10\)
\(\Rightarrow\left(x-1\right);\left(8-x\right)\in\left\{1;2;5;10\right\}\)
ta có
x-1 | 1 | 2 | 5 | 10 |
8-x | 10 | 5 | 2 | 1 |
x | (s) | 3 | 6 | (s) |
từ bảng trên cho ta thấy
x chỉ có thể lả 3 hoặc 6
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\(-4:\frac{1}{3}\left(\frac{1}{2}-\frac{1}{6}\right)< n< \frac{-2}{3}\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
\(\Rightarrow-4\cdot3\left(\frac{3}{6}-\frac{1}{6}\right)< n< -\frac{2}{3}\left(\frac{4}{12}-\frac{6}{12}-\frac{9}{12}\right)\)
\(\Rightarrow-4\cdot3\cdot\frac{1}{3}< n< -\frac{2}{3}\cdot\left(-\frac{11}{12}\right)\)
\(\Rightarrow-4< n< -\frac{1}{3}\cdot\left(-\frac{11}{6}\right)=\frac{11}{18}\)
=> \(-4< n< \frac{11}{18}\)
=> \(-\frac{72}{18}< n< \frac{11}{18}\)
Đến đây bạn tự xét đi nhé
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Bài giải
\(\frac{3}{4}-\left[\left(\frac{-5}{3}-\left(\frac{1}{12}+\frac{2}{9}\right)\right)\right]\)
\(=\frac{3}{4}+\frac{71}{36}\)
\(=\frac{49}{18}\)
\(\frac{3}{4}-\left[\left(\frac{-5}{3}\right)-\left(\frac{1}{12}+\frac{2}{9}\right)\right]\)
\(=\frac{3}{4}-\left[\left(\frac{-5}{3}\right)-\frac{11}{36}\right]\)
\(=\frac{3}{4}-\left(\frac{-71}{36}\right)\)
\(=\frac{49}{18}\)
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\(2014:\left(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1\frac{2}{5}-\frac{7}{9}+\frac{7}{11}}\cdot\frac{1\frac{1}{6}+0,875-0,7}{\frac{1}{3}+0,25-\frac{1}{5}}\right)\)
\(=2014:\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}\cdot\frac{\frac{7}{6}+\frac{7}{8}-\frac{7}{10}}{\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}\right)\)
\(=2014:\left(\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}\cdot\frac{\frac{7}{6}+\frac{7}{8}-\frac{7}{10}}{\frac{2}{6}+\frac{2}{8}-\frac{2}{10}}\right)\)
\(=2014:\left(\frac{2}{7}\cdot\frac{7\left(\frac{1}{6}+\frac{1}{8}-\frac{1}{10}\right)}{2\left(\frac{1}{6}+\frac{1}{8}-\frac{1}{10}\right)}\right)\)
\(=2014:\left(\frac{2}{7}\cdot\frac{7}{2}\right)=2014\)
(23 : 4). 2x+1 = 64
=> (8 : 4) . 2x+1 = 64
=> 2.2x+1 = 64
=> 2x+1 = 32
=> 2x+1 = 25
=> x + 1 = 5 => x = 4
Vậy x = 4
\(\left(2^3\div4\right).2^{\left(x+1\right)}=64\Leftrightarrow\left(8\div4\right).2^x.2^1=2^6\)
\(\Leftrightarrow2^1.2^x.2^1=2^2.2^x=2^6\Leftrightarrow2^{\left(2+x\right)}=2^6\)
\(\Leftrightarrow2+x=6\Leftrightarrow x=6-2=4\)
Vậy \(x=4\)